| Exam Board | Edexcel |
|---|---|
| Module | C1 (Core Mathematics 1) |
| Year | 2010 |
| Session | January |
| Marks | 8 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Tangents, normals and gradients |
| Type | Find tangent at given point (polynomial/algebraic) |
| Difficulty | Moderate -0.8 This is a straightforward C1 differentiation question requiring quotient rule or algebraic simplification followed by basic differentiation, then finding a tangent equation using standard y - y₁ = m(x - x₁). Both parts are routine textbook exercises with no problem-solving insight needed, making it easier than average but not trivial due to the algebraic manipulation required. |
| Spec | 1.07i Differentiate x^n: for rational n and sums1.07m Tangents and normals: gradient and equations |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| \(y = \frac{x^2 - 5x - 24}{x} = x - 5 - 24x^{-1}\) | M1 A1 | Multiply out to get \(x^2 + bx + c\), \(b\neq 0\), \(c\neq 0\), and divide by \(x\) (not \(x^2\)). One correct term e.g. \(x\)... is sufficient evidence of division attempt |
| \(\frac{dy}{dx} = 1 + 24x^{-2}\) or \(\frac{dy}{dx} = 1 + \frac{24}{x^2}\) | M1 A1 | Dependent on 1st M: evidence of \(x^n \to kx^{n-1}\) for one \(x\) term (not just constant term). Mark not given if numerator and denominator differentiated separately |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| \(x = 2\): \(y = -15\) | B1 | Allow if seen in part (a) |
| \(\frac{dy}{dx} = 1 + \frac{24}{4} = 7\) | B1ft | Follow-through from candidate's non-constant \(\frac{dy}{dx}\). Must be simplified to a "single value" |
| \(y + 15 = 7(x-2)\) (or equiv., e.g. \(y = 7x - 29\)) | M1 A1 | Equation in any form of straight line through \((2, -15)\) with candidate's \(\frac{dy}{dx}\) value as gradient. Note: \(y-(-15) = 7(x-2)\) is A0 (unresolved minus minus) |
## Question 6:
**(a)**
| Answer/Working | Marks | Guidance |
|---|---|---|
| $y = \frac{x^2 - 5x - 24}{x} = x - 5 - 24x^{-1}$ | M1 A1 | Multiply out to get $x^2 + bx + c$, $b\neq 0$, $c\neq 0$, and divide by $x$ (not $x^2$). One correct term e.g. $x$... is sufficient evidence of division attempt |
| $\frac{dy}{dx} = 1 + 24x^{-2}$ or $\frac{dy}{dx} = 1 + \frac{24}{x^2}$ | M1 A1 | Dependent on 1st M: evidence of $x^n \to kx^{n-1}$ for one $x$ term (not just constant term). Mark not given if numerator and denominator differentiated separately |
**(b)**
| Answer/Working | Marks | Guidance |
|---|---|---|
| $x = 2$: $y = -15$ | B1 | Allow if seen in part (a) |
| $\frac{dy}{dx} = 1 + \frac{24}{4} = 7$ | B1ft | Follow-through from candidate's non-constant $\frac{dy}{dx}$. Must be simplified to a "single value" |
| $y + 15 = 7(x-2)$ (or equiv., e.g. $y = 7x - 29$) | M1 A1 | Equation in any form of straight line through $(2, -15)$ with candidate's $\frac{dy}{dx}$ value as gradient. Note: $y-(-15) = 7(x-2)$ is A0 (unresolved minus minus) |
---6. The curve $C$ has equation
$$y = \frac { ( x + 3 ) ( x - 8 ) } { x } , \quad x > 0$$
\begin{enumerate}[label=(\alph*)]
\item Find $\frac { \mathrm { d } y } { \mathrm {~d} x }$ in its simplest form.
\item Find an equation of the tangent to $C$ at the point where $x = 2$
\end{enumerate}
\hfill \mbox{\textit{Edexcel C1 2010 Q6 [8]}}